771 reputation
26
bio website harlanjbrothers.com
location Connecticut, USA
age
visits member for 1 year, 4 months
seen Oct 17 at 5:41

I'm an inventor, mathematician, composer, and educator. I take particular interest in fractal geometry and number theory. Mathematica is one of my primary research tools.


Sep
24
awarded  Autobiographer
Jun
15
awarded  Yearling
Jun
10
answered What is an example of real application of cubic equations?
Feb
1
comment Surprising identities / equations
Nice, @FredKline. The above link leads to a list of formulas at: brotherstechnology.com/math/e-formulas.html. This is a variation of Formula (26) with n=1: \begin{equation} 1=\sum _{k=1}^{\infty } \frac{k}{(k+1)!}=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+\frac{5}{6!‌​}+\frac{6}{7!}+\ldots~. \end{equation}
Jan
8
revised Surprising identities / equations
More examples, typo
Jan
8
answered Surprising identities / equations
Sep
18
revised How can one prove that $e<\pi$?
Attempted to address the question more specifically
Sep
18
answered How can one prove that $e<\pi$?
Sep
15
revised What are some examples of mathematics that had unintended useful applications much later?
Fixed grammar
Sep
15
revised What are some examples of mathematics that had unintended useful applications much later?
Corrected typo
Sep
10
answered What are some examples of mathematics that had unintended useful applications much later?
Aug
7
comment Is The *Mona Lisa* in the complement of the Mandelbrot set.
It appears you are envisioning a two-dimensional analog of the idea that eventually one can find any arbitrary sequence of digits in an infinite sequence with a uniform random distribution. The question, however, does not seem clearly formed. Most importantly, if the escape values in your array are unique, how can they generate the Mona Lisa whose array must contain identical values? If, as you see it, the Mona Lisa's array does not contain identical values, then this is akin to claiming that, for example, a 1000 x 1000 array of the numbers 1 to 10,000 is equivalent to the Mona Lisa.
Aug
5
revised $\sum_{k=0}^{n/2} {n\choose{2k}}=\sum_{k=1}^{n/2} {n\choose{2k-1}}$, Combinatorial Proof:
Improved language
Aug
5
revised $\sum_{k=0}^{n/2} {n\choose{2k}}=\sum_{k=1}^{n/2} {n\choose{2k-1}}$, Combinatorial Proof:
Original question was clarified - edited answer accordingly.
Aug
5
answered $\sum_{k=0}^{n/2} {n\choose{2k}}=\sum_{k=1}^{n/2} {n\choose{2k-1}}$, Combinatorial Proof:
Aug
1
comment Can the golden ratio accurately be expressed in terms of e and $\pi$
@A.Rex Yes. I didn't get to go into a properly detailed response yesterday, but because the partial numerators of the CF decrease exponentially while the partial denominators remain constant, the convergence is rapid. Just keeping the $e^{-2\pi}$ term gives 8 correct decimal places. The convergent for the next term gives 16 d.p.a., followed by 27, 41, 57, and 76 d.p.a. if we go as far as $e^{-12\pi}$. ...Not that anyone would actually want to use this circuitous approach. :)
Jul
30
awarded  Supporter
Jul
29
awarded  Good Answer
Jul
29
awarded  Mortarboard
Jul
29
awarded  Nice Answer